\cos 2 A \cos 4 A \ldots \ldots \cos 2^{n-1} A=\frac{\sin 2^{n} A}{2^{

1.	\cos 2 A \cos 4 A \ldots \ldots \cos 2^{n-1} A=\frac{\sin 2^{n} A}{2^{n} \sin 2 A}\sin ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{5 \pi}{8}+\sin ^{4} \frac{1 \pi}{8}
Answer» sin⁴(π/8) + sin⁴(3π/8) + sin⁴(5π/8) + sin⁴(7π/8) = [sin²(π/8)]² + [sin²(3π/8)]² + [sin²(5π/8)]² + [sin²(7π/8)]² = [{1 - cos(π/4)}/2]² + [{1 - cos(3π/4)}/2]² + [{1 - cos(5π/4)}/2]² + [{1 - cos(7π/4)}/2]² = [{1 - √2/2 }/2]² + [{1 + √2/2 }/2]² + [{1 + √2/2}/2]² + [{1 - √2/2}/2]² = 2[{1 - √2/2 }/2]² + 2[{1 + √2/2 }/2]² = 2[{2 - √2}/4]² + 2[{2 + √2}/4]² = (1/8)(2 - √2)² + (1/8)(2 + √2)² = (1/8)[(4 - 4√2 + 2) + (4 + 4√2 + 2)] = 12/8 = 3/2

\cos 2 A \cos 4 A \ldots \ldots \cos 2^{n-1} A=\frac{\sin 2^{n} A}{2^{n} \sin 2 A}\sin ^{4} \frac{\pi}{8}+\sin ^{4} \frac{3 \pi}{8}+\sin ^{4} \frac{5 \pi}{8}+\sin ^{4} \frac{1 \pi}{8}

Answer»

sin⁴(π/8) + sin⁴(3π/8) + sin⁴(5π/8) + sin⁴(7π/8)

= [sin²(π/8)]² + [sin²(3π/8)]² + [sin²(5π/8)]² + [sin²(7π/8)]²

= [{1 - cos(π/4)}/2]² + [{1 - cos(3π/4)}/2]² + [{1 - cos(5π/4)}/2]² + [{1 - cos(7π/4)}/2]²

= [{1 - √2/2 }/2]² + [{1 + √2/2 }/2]² + [{1 + √2/2}/2]² + [{1 - √2/2}/2]²

= 2[{1 - √2/2 }/2]² + 2[{1 + √2/2 }/2]²

= 2[{2 - √2}/4]² + 2[{2 + √2}/4]²

= (1/8)(2 - √2)² + (1/8)(2 + √2)²

= (1/8)[(4 - 4√2 + 2) + (4 + 4√2 + 2)]

= 12/8 = 3/2